Answer:
C.
Step-by-step explanation:
p(x)=sin(x) is an odd function since sin(-x)=-sin(x).
q(x)=cos(x) is an even function since cos(-x)=cos(x).
r(x)=tan(x) is an odd function since tan(-x)=-tan(x).
s(x)=csc(x) is an odd function since csc(-x)=-csc(x).
So the only contender seems to be C.
Let's check. To check we have to plug in (-x) in place of (x) and see if we get the same function back since we are looking for an even function.
Replace (x) with (-x):
since cosine is even; that is cos(-u)=cos(u) where u in this case is .
So f is even.
14 Hamburgers because if you multiply 38 by 2 because it was hamburgers AND cheesburgers, you would get 76...then subtract 62 from 76 and you get 14
Hello from MrBillDoesMath!
Answer:
40
Discussion:
4+4+4+4+4+4+4+4+4+4 (*)
There are 10 4's in (*) so
(*) = 4 * 10 = 40
Thank you,
MrB
Answer:
atupagin mo module mo wag puro brainly
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.